I need a physics genius

I have been working on this problem and I cant get the right answer for the second part, I keep on getting 8.75 and that wrong. can someone answer this?

The linear acceleration of a particle is given by a(t) = 7.00 − t^3 where the acceleration is in m/s^2 and the time is in seconds. At t = 0 the position of the particle is 5.00 meters, and at t = 2.00 seconds, the particle's velocity is 8.00 m/s. (a) What is the velocity of the particle at t = 6.00 s? (b) What is the position of the particle at t = 1.00 s?

integrate twice;
v = INT[adt] = 7t - (1/4)t^4 + b
x = INT[vdt] = (7/2)t^3 - (1/20)t^5 + bt + c

Where "b" & "c" are constants of integration. You find them by applying the given info.
at t=2 ,v=8
8 = 7(2) - (1/4)(2)^4 + b
b = -2

at t=0 x=5
5 = 0 + 0 + 0 + c
c = 5

Now put these values for "b" & "c" back into the eqs for "v" & "x";
v = 7t - (1/4)t^4 - 2
x = (7/2)t^2 - (1/20)t^5 - 2t + 5

Now you can find "x" & "v" for any time you want.

a(t) = d2x/dt2 = 7 − t^3
Integrate
v = dx/dt = Integral(7 − t^3) = 7t - t^4/4 + C

When t=2s, v = 8m/s so:
8 = 7x2 - 2^4/4 + C
C = 8 - 10
= -2

v = 7t - t^4/4 - 2

When t = 6s, v = 7x6 -6^4/4 - 2
v = -284m/s
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b) v = dx/dt = 7t - t^4/4 - 2
x = integral (7t - t^4/4 - 2)
x= 7t^2/2 - t^5/20 - 2t + C

When t= 0, x = 5 so CX = 5 giving
x= 7t^2/2 - t^5/20 - 2t + 5

When t=1s, x = 7/2 - 1/20 -2 + 5
x = 6.45m